Match List-I with List-II -

List-I (Separation of)	List-II (Separation Technique)
(A) Aniline from aniline-water mixture	(I) Simple distillation
(B) Glycerol from spent-lye in soap industry	(II) Fractional distillation
(C) Different fractions of crude oil in petroleum industry	(III) Distillation at reduced pressure
(D) Chloroform-Aniline mixture	(IV) Steam distillation

Choose the correct answer from the options given below:

### Core Logic To determine the IUPAC name of the compound shown in

: 1. Identify the principal functional group, which is the double bond (alkene). 2. Find the longest carbon chain containing the double bond: mathrmC1(H_2Br) - C2(CH_3) = C3(H) - C4(H_3)$$\mathrm{C1(H_2Br) - C2(CH_3) = C3(H) - C4(H_3)}$$ The longest chain has 4$4$ carbons, which means the parent alkane is butane, and with a double bond it's "but-2-ene". 3. Number the chain from the end that gives lower locants to the double bond. Starting from left or right both give the double bond at position 2$2$. However, starting from left gives substituent locants as 1$1$ (for bromo) and 2$2$ (for methyl), whereas starting from right gives substituent locants as 3$3$ and 4$4$. 4. Hence, correct numbering is: - textC1$\text{C1}$: bonded to Bromine (-textBr$-\text{Br}$) - textC2$\text{C2}$: bonded to Methyl (-textCH_3$-\text{CH}_3$) - textC3$\text{C3}$: alkene carbon - textC4$\text{C4}$: terminal methyl group

Combining these rules, the name is: **1-Bromo-2-methylbut-2-ene**. ### Pattern Recognition Double bond takes precedence over halogen substituent in numbering direction. If double bond is symmetrical (at position 2$2$ in a 4$4$-carbon chain), use the substituent positions to break the tie, choosing lowest possible locants (1$1$ and 2$2$ vs 3$3$ and 4$4$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula \% mathrmC = frac1244 times fractextMass of mathrmCO_2 text producedtextMass of organic compound taken times 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{\text{Mass of } \mathrm{CO}_2 \text{ produced}}{\text{Mass of organic compound taken}} \times 100$$ ### Core Logic Given: - Mass of organic compound taken = 500 text mg = 500 times 10^-3 text g$= 500 \text{ mg} = 500 \times 10^{-3} \text{ g}$ - Mass of mathrmCO_2$\mathrm{CO}_2$ produced = 220 text mg = 220 times 10^-3 text g$= 220 \text{ mg} = 220 \times 10^{-3} \text{ g}$ Using the formula: \% mathrmC = frac1244 times frac220 times 10^-3500 times 10^-3 times 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{220 \times 10^{-3}}{500 \times 10^{-3}} \times 100$$ \% mathrmC = frac1244 times frac220500 times 100$$\% \mathrm{C} = \frac{12}{44} \times \frac{220}{500} \times 100$$ \% mathrmC = frac1244 times 44 = 12 \%$$\% \mathrm{C} = \frac{12}{44} \times 44 = 12 \%$$ Thus, the percentage of carbon is 12$12$. ### Pattern Recognition Carbon dioxide has exactly 12/44 approx 27.27\%$12/44 \approx 27.27\%$ carbon by mass. Multiply the mass fraction of mathrmCO_2$\mathrm{CO}_2$ (220/500 = 0.44$220/500 = 0.44$) by 12/44$12/44$ to directly get 0.12$0.12$ or 12\%$12\%$. ### Evaluation Rubric / Model Answer A perfect step-by-step conversion of organic compound mass and combustion carbon dioxide mass to obtain a precise 12$12$ percent carbon composition. ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let us apply official IUPAC priority indexing rules: 1. **Principal Functional Group**: The hydroxyl group (-textOH$-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the -textOH$-\text{OH}$ group is assigned position **C-1**. 2. **Numbering Direction**: We must number through the ring towards the double bond to assign it the lowest possible locant. Hence, the alkene carbons are given coordinates **C-2** and **C-3**. 3. **Locating Side Chains**: Proceeding with this direction puts the ethyl group at position **C-4**.

Assembling the structural parts alphabetically: * Substituent: `4-Ethyl` * Parent root: `cyclopent-2-en` * Suffix: `1-ol` Combined IUPAC format: **4-Ethylcyclopent-2-en-1-ol**. ### Pattern Recognition Principal suffix priority hierarchy: -textOH > textDouble bond > textAlkyl side-chain$-\text{OH} > \text{Double bond} > \text{Alkyl side-chain}$. Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let us analyze each term carefully: * **A. Carbocation**: Features a positively charged trivalent carbon atom. It represents an sp^2$sp^2$ hybridized carbon with an empty unhybridized p-orbital.

* **B. Carbon Free Radical**: Contains a trivalent carbon carrying a single unpaired lone electron. It typically exhibits sp^2$sp^2$ or sp^3$sp^3$ hybridization depending on structural environments.

* **C. Nucleophile**: An electron-rich chemical species containing a lone pair or negative charge capable of donating/supplying a pair of electrons. * **D. Electrophile**: An electron-deficient chemical species possessing empty low-lying orbitals capable of accepting/receiving a pair of electrons. ### Step 1: Alignment Matrix Matching each item yields: * textA rightarrow textIII$\text{A} \rightarrow \text{III}$ * textB rightarrow textIV$\text{B} \rightarrow \text{IV}$ * textC rightarrow textI$\text{C} \rightarrow \text{I}$ * textD rightarrow textII$\text{D} \rightarrow \text{II}$ This sequence aligns flawlessly with Option (4). ### Pattern Recognition Nucleophiles donate ('nucleo-loving' = seeks positive sites with its electrons), Electrophiles accept ('electro-loving' = seeks electron density). Carbocations explicitly harbor a vacant p-orbital because of their positive charge configuration, making identification extremely swift. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Percentage of Hydrogen in organic analysis: \%textH = frac218 times fractextMass of H_2OtextMass of Compound times 100$$\%\text{H} = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of Compound}} \times 100$$ Percentage of Carbon: \%textC = frac1244 times fractextMass of CO_2textMass of Compound times 100$$\%\text{C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of Compound}} \times 100$$ Percentage of Oxygen: \%textO = 100 - (\%textC + \%textH)$$\%\text{O} = 100 - (\%\text{C} + \%\text{H})$$ ### Execution Step 1: Compute the mass percent of Hydrogen: \%textH = frac218 times frac0.1270.210 times 100 = frac0.2543.78 approx 6.72\%$$\%\text{H} = \frac{2}{18} \times \frac{0.127}{0.210} \times 100 = \frac{0.254}{3.78} \approx 6.72\%$$ Step 2: Compute the mass percent of Carbon: \%textC = frac1244 times frac0.3070.210 times 100 = frac3.6849.24 approx 39.87\%$$\%\text{C} = \frac{12}{44} \times \frac{0.307}{0.210} \times 100 = \frac{3.684}{9.24} \approx 39.87\%$$ Step 3: Deduce the remaining mass percent of Oxygen: \%textO = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$$\%\text{O} = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$$ Thus, the values of hydrogen and oxygen percentage are 6.72\%$6.72\%$ and 53.41\%$53.41\%$, matches with Option (2). ### Pattern Recognition Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques